Excel in math and science.

Join using Facebook Join using Google Join using email

Forgot password? New user? Sign up

Existing user? Log in

Your answer seems reasonable. Find out if you're right!

That seems reasonable. Find out if you're right!

Join using Facebook Join using Google Join using email

Forgot password? New user? Sign up

Existing user? Log in

Classical Mechanics

Damped Oscillators

Concept Quizzes

Exponential decay from damping forces
Damping and amplitude decreases
Damped Oscillators - Problem Solving

Damped Oscillators - Problem Solving

A body of mass $m = 1 \text{ kg}$ is oscillating on a spring with spring constant $k = 2 \text{ N/m}.$ There is a friction proportional to the body's velocity, represented by the function $f = -bv = -2v \text{ (N)}$ . Initially, the position of the body was $x(t=0) = 2 \text{ m }$ and the velocity was $v(t=0) = 0 \text{ m/s },$ where $x$ denotes the difference in the length of the spring from its original length. What is the amplitude of the body when $t = 2 \text{ s } ?$

Assume that $\sqrt{2} = 1.5$

3\exp(-2) \text{ (m)}

3\exp(-4) \text{ (m)}

4\exp(-2) \text{ (m)}

3\exp(-3) \text{ (m)}

Show explanation

View wiki

by Brilliant Staff

A body of mass $m = 2 \text{ kg}$ is oscillating on a spring with spring constant $k = 4 \text{ N/m}.$ There is a friction proportional to the body's velocity, represented by the function $f = -bv = -4v \text{ (N)}$ . Initially, the position of the body was $x(t=0) = 5 \text{ m }$ and the velocity was $v(t=0) = 0 \text{ m/s },$ where $x$ denotes the difference in the length of the spring from its original length. What is the velocity of the body when $t = \frac{\pi}{2} \text{ s } ?$

{ -9\exp\left(-\frac{\pi}{2}\right) \text{ (m/s)}}

{ -10\exp\left(-\frac{\pi}{2}\right) \text{ (m/s)}}

{ -11\exp\left(-\frac{\pi}{2}\right) \text{ (m/s)}}

{ -15\exp\left(-\frac{\pi}{2}\right) \text{ (m/s)}}

Show explanation

View wiki

by Brilliant Staff

A body of mass $m = 3 \text{ kg}$ is oscillating on a spring with spring constant $k = 3 \text{ N/m}.$ There is a friction proportional to the body's velocity, represented by the function $f = -bv = -6v \text{ (N)}.$ Initially, the position of the body was $x(t=0) = 1 \text{ m }$ and the velocity was $v(t=0) = 0 \text{ m/s },$ where $x$ denotes the difference in the length of the spring from its original length. What is the velocity of the body when $t = 1 \text{ s } ?$

-1\exp(-3) \text{ (m/s)}

-2\exp(-1) \text{ (m/s)}

-1\exp(-1) \text{ (m/s)}

-1\exp(-2) \text{ (m/s)}

Show explanation

View wiki

by Brilliant Staff

A body of mass $m = 6 \text{ kg}$ is oscillating on a spring with spring constant $k = 12 \text{ N/m}.$ There is a friction proportional to the body's velocity, represented by the function $f = -bv = -12v \text{ (N)}.$ Initially, the position of the body was $x(t=0) = 4 \text{ m }$ and the velocity was $v(t=0) = 0 \text{ m/s },$ where $x$ denotes the difference in the length of the spring from its original length. What is the ratio of the amplitude to the initial amplitude when $t = 5 \text{ s } ?$

\exp(-5)

\exp(-8)

\exp(-7)

\exp(-6)

Show explanation

View wiki

by Brilliant Staff

A body of mass $m = 3 \text{ kg}$ is oscillating on a spring with spring constant $k = 3 \text{ N/m}.$ There is a friction proportional to the body's velocity, represented by the function $f = -bv = -6v \text{ (N)}$ . Initially, the position of the body was $x(t=0) = 3 \text{ m }$ and the velocity was $v(t=0) = 0 \text{ m/s },$ where $x$ denotes the difference in the length of the spring from its original length. What is the position of the body when $t = 3 \text{ s } ?$

12\exp(-3) \text{ (m)}

12\exp(-5) \text{ (m)}

13\exp(-3) \text{ (m)}

12\exp(-4) \text{ (m)}

Show explanation

View wiki

by Brilliant Staff